4 Answers
4

We are told that $\Pr(B|A)\gt \Pr(B)$. So knowing that $A$ holds makes $B$ more likely. It stands to reason that knowing that $A'$ holds makes $B$ less likely. "It stands to reason" is presumably not enough for a proof, so we give full details.

Of course we will use $\Pr(B|A)=\frac{\Pr(A\cap B)}{\Pr(A)}$. For this we need $\Pr(A)\ne 0$. From $\Pr(B|A)\gt \Pr(B)$ we get
$$\Pr(A\cap B)\gt \Pr(A)\Pr(B).\tag{$1$}$$

Note that
$$\Pr(A\cap B)+\Pr(A'\cap B)=\Pr(B).$$
Substituting in Equation $(1)$ we get
$$\Pr(B)-\Pr(A'\cap B)\gt \Pr(A)\Pr(B),$$
which can be rewritten as
$$\Pr(A'\cap B)\lt \Pr(B)-\Pr(A)\Pr(B)=\Pr(B)(1-\Pr(A))=\Pr(B)\Pr(A').$$
If $\Pr(A')\ne 0$, we can divide, obtaining
$$\frac{\Pr(A'\cap B)}{\Pr(A')}\lt \Pr(B).$$
This says exactly what we want, since the left side is just $\Pr(B|A')$.

Intuitively this is the statement that "the weighted average of two numbers must lie between the two numbers". $P(B)$, the probability of $B$ happening over the whole sample space, is the weighted average of $P(B|A)$, the probability of $B$ happening over the sample space for which $A$ happens, and $P(B|A')$, the probability of $B$ happening over the remaining part of the sample space. Hence we must have that if $P(B|A) > P(B)$, the left over bit satisfies $P(B) > P(B|A')$.